 Welcome back to our lecture series math 1210 calculus one for students at Southern Utah University as usual of your professor today Dr. Angela misledine in this lecture number 14 I want to talk about limits which approach infinity for which there's really two types of limits We have in mind So one type of limit we mean as you approach infinity would be something like When the variable x itself approaches plus or minus infinity of f of x what happens in that type of setting? Well, this leads to the notion of in behavior. Sometimes it leads to something like a horizontal acetote That's something we'll talk about in later videos in this lecture The the type of limits as that approach infinity that I want to talk about in this Video will be limits of the form where we take the limit as x approaches a from the right or the left So a is itself a finite number and it turns out that this limit is equal to positive or negative infinity So the limit itself is Infinite Which is a different type of limit at infinity I would like to talk about and so in this discussion This leads to the notion of a vertical acetote if a left-handed or right-handed limit approaches positive or negative infinity This is manifested as a vertical acetote on the graph of our function and typically that Function will have or that limit. I should say we'll have the form of r divided by 0 where r is any Non-zero real number and so if you get something like a non-zero divided by a zero That's going to give you a limit of plus or minus infinity And that'll indicate you have a vertical acetote on your graph So we I want to first mention some functions that we've talked about before that do have Vertical acentotes one famous family of functions would be the reciprocal functions, of course So if we think of our x and y axis here for a moment if we take the function So like 1 over x you're gonna get a graph that looks something like this In the first quadrant it would look something like this in the In the third quadrant there. So this would be like y equals 1 over x I said to the first but honestly if you have any odd power Your function is gonna look something like this and some things I can say about the vertical acentotes of this function This reciprocal function. It'll have a vertical acentote at x equals 0 aka the y-axis For which you can see that on our picture right here You have this vertical acentote, but we can actually do a little bit better than that Notice that as we approach zero from the right and the left we can say something about that So as you approach zero from the right Then your function 1 over x to an odd power is going to give you positive infinity on the other hand if we take the limit as x approaches zero from the left of 1 over x to some odd power This is actually going to give us negative infinity as we can see it's going off towards negative infinity there If we were trying to graph this for even functions Let's say we looked at 1 over x squared 1 over x to the fourth etc If we wanted to graph this for even functions For which I think I'm just going to rewrite it To avoid any potential confusion there. We take 1 over x to some even power Then some things we can see about the graph will be very similar some things are going to be very different We're gonna see that for an even powered reciprocal function You will still have a vertical acentote at the y-axis The first quadrant is actually going to look very similar and therefore you see the same limit the limit as x approaches zero from the right of 1 over x to an even power This is still going to equal infinity The main difference is what happens to the left side of the y-axis You're going to get something that's actually the mirror image of what you saw in the first quadrant And therefore you're going to see a mirror image of the limits You're going to see that as x approaches zero from the left of 1 over x to an even power This is also going to turn out to be Positive infinity and that's what this information right here is trying to summarize here that whether your exponent is even or odd If you approach zero from the right 1 over x to the end will become infinite On the other hand if you approach zero from the left 1 over x to the end will approach either positive infinity or negative infinity positive infinity when you're even and Negative infinity when you're odd and this has to do with of course Powers of negative numbers notice if you take negative one squared you're going to get a positive one But if you take negative one cubed you're going to get a negative one even powers of a negative will give you something positive so taking an even power of Numbers that are negative but getting close to zero will will it make it positive the answer you get positive infinity on the other hand if you Take an odd power of negative numbers as they're getting closer to infinity That's going to always be negative and that gets you towards negative infinity here So while you can memorize these statements I think it's better to think about the graphs of these things if you remember the graphs of these reciprocal functions You can always reverse engineer these statements about limits as you're approaching the vertical asymptote of our function or not And so this becomes even more true. I think when we start talking about vertical asymptotes of Trigometric functions so while sine and cosine don't have any vertical asymptotes the other four trigonometric functions do so imagine tangent for example tangent if you approach Pi halves plus pi k where k is any integer from the right You're going to get negative infinity But if you approach pi halves plus pi k from the left then you're going to get positive infinity that can be very difficult to forget What's going on there? It's like positive negative You know, it's almost like a random guess that you're gonna be right half the time but also wrong half the time when it comes to tangent I think it's useful if you want to think about it graphically To think about the principal branch of tangent, right? So tangent has is its principal branch this continuous piece right here, but then when you hit pi on The right or if you hit neg excuse me if you hit pi halves on the right x equals pi halves Or if you hit negative pi halves on the left this is where you're going to be Hitting your vertical asymptotes and so this is where we can ask or make statements about limits So what I would like you to think of is the following Okay, if you're taking the limit as x approaches pi halves from the left of Tangent here you can see that we're going to be approaching positive infinity as tangent approach a vertical asymptote from the left It'll go off towards positive Infinity but as we approach it from the right We see that the limit as x approaches negative pi halves from the left of tangent of x We're gonna end up with negative infinity Because of this picture right here and therefore because tangents periodic if I were to draw the picture again I know though if I were to approach if I was to approach pi halves from the right That's gonna be negative infinity. I don't necessarily memorize these limit statements. Oh if you approach from the right It's negative infinity if you approach from the left is infinity what I would say is remember the picture Because if you can see the picture you can always answer the question Okay, let's see if tangent is approaching a vertical asymptote from the left It's goes it's gonna be positive from the right to be negative. You can then figure out these things Similar statements can be said for cotangent for secant. It gets a little bit more complicated There's actually four possibilities to consider for secant cotangents a lot easier Because with cotangent let me kind of erase this Do-do-do-do cotangent your picture is gonna look like so Your asymptotes in our slightly different location. You're gonna get your asymptotes at zero and at pi Pi and zero so if ever you're taking the limit as X approaches any any vertical asymptote, right? So this is gonna be multiples of pi if you approach it from the left of cotangent You're gonna end up with a negative infinity, but if you're approaching from the right Zex approaches pi k from the right of cotangent of X. This is gonna be positive infinity You can remember this from the graph, which is what we're saying down here as well Let's try this one more time with say secant What is your basic secant function look like you're gonna get something looks like this and you're gonna get something Looks like this again. These are not the best the best drawn secant functions I do apologize for that But the intuition is what matters here not the artistic prowess, right? You see that it depends on which asymptote you're at so it gets a little it gets much more tricky with secant, right? So secant as the reciprocal of cosine Secant will be undefined. I'll have a vertical asymptote whenever cosine goes to zero which is gonna happen at pi halves It would happen at three pi halves. It would happen a negative pi halves So the exact same locations that tangent as a vertical asymptote and so this is where we have to be careful So if we're approaching pi halves from the left Why is going to approach infinity? But if we're approaching pi halves from the right Then why will approach negative infinity? It becomes almost impossible to just purely memorize all these things these limit statements That is if you take all of these limit statements and their greatest generality It becomes very difficult and so I would suggest when it comes to trigonometric functions rely on the graphs If I can think of what the graph of secant looks like remembering the vertical asymptotes of secant and tangent Will be exactly when cosine goes to zero I bet you I can reverse engineer this thing and another thing to remember about secant's graph is as it's the reciprocal of cosine You kind of see this like bending away From the cosine curve you can reproduce the picture secant on Demand same thing with cosecant and thus you could calculate these limits on demand when really is necessary the other family of functions in In a in a in a calculus setting that we really would want to discuss that have vertical asymptotes are gonna be logarithms The natural log is a good candidate to explain what's going on here for your standard natural for your standard logarithm You're gonna get a graph that looks something like the following Right why equals the natural log of x of course if you modify the base That'll stretch this thing this will make this thing taller or smaller can also reflect and you get something that looks like the following In either case though, you will have a vertical asymptote at The y-axis again x equals zero and so for the natural log We see that as x approaches zero from the right the natural log will go towards negative infinity It's going down down down down down down down down down Again, if you reflected this natural log Or any logarithm across the x-axis you can't get a decreasing logarithm like so in this situation Then as you go towards zero from the right you'd be approaching positive infinity So rely on your knowledge of the graphs the reason Honestly, I think the main reason why it's so important for you as a calculus student to be able to represent functions in different ways Is that different representations have advantages? If you tried to do limit calculations only algebraically that'll only get you so far There are times where a graphical approach will be superior and I think when it comes to discussing vertical asymptotes It's very important to remember the graphs of the functions that are in play Let's look at a few examples and see what's going on here So if I have the function or if I had to compute the limit excuse me the limit as x approaches 2 of 3x minus 2 over x minus 2 The first thing I would be want to do is plug in x equals 2, right? because When because it's a rational function, it's continuous on its domain So as long as the denominator doesn't go to zero then the limit calculation will just be the function evaluation So I'm just gonna plug things in to see what happens You're gonna get 3 times 2 minus 2 over 2 minus 2 for which the denominator We can very quickly see becomes 0 that gives us some concern But if the numerator is also 0 it means that I might be able to simplify the rational function to determine what the limit is I have a removable discontinuity But we see that the numerator is not that it's not gonna be 0 here We get 3 times 2 which is 6 minus 2 which is 4 we get 4 Over 0 which this indicates we don't have a removable point, but this does tell us we have a vertical asymptote So if we're looking If we're looking for the vertical asymptotes of the function then it's exactly these limits of the form a non-zero Over zero so we found a vertical asymptote But maybe we want to find a little bit more about what's going on here So if we think about it graphically Right, what's our function doing? Well, is it maybe both sides of 2 go off towards infinity? Maybe one side goes to infinity the other side goes to negative infinity Maybe they both go down to negative infinity or maybe the right side goes to negative infinity the left side goes to positive infinity We have to investigate this thing a little bit more. Okay, because there are these different possibilities so if we consider the limit as x approaches 2 from the right of 3x minus 2 over x minus 2 well We're gonna do here. I'm gonna suggest a strap I'm gonna suggest a strategy to you that is not exactly in common use, but works super well Well, we're gonna do it's since we're approaching 2 from the right We're gonna plug in what we will denote as 2 plus So we're gonna get 2 plus minus 2 over here. All right, and we can do that in the numerator as well It's not actually necessary in the numerator, but we'll just do it for the sake of it Okay, and so what I want you to think about this is that when I say 2 plus I want you to think about it is what if we were to pick a number that's a little bit bigger than 2 All right, if you take a number that's a little bit bigger than 2 and you times it by 3 You're gonna get something that's a little bit bigger than 6. Okay, and you're gonna subtract 2 from that Well, if you take a number that's a little bit bigger than 6 and you subtract 2 from it You're gonna get a number that's a little bit bigger than 4 Okay. On the other hand though, if you take a number that's a little bit bigger than two and you subtract from it two, you're going to get something that's a little bit bigger than zero. Okay. And so you get zero, you get four plus over zero plus. But in all reality, when it comes to a number that's a little bit bigger than four versus a number a little bit smaller than four, they're both going to be positive. So I can actually drop the plus on the four. I'm just going to write four over zero plus. So why is it so significant for the zero here? Well, the fact that we have this form of four over zero, we know that we have a vertical asymptote. So these left and right hand limits, they're going to be either plus or minus infinity. It's the sign that we're trying to figure out. We know it's infinite. We don't know the sign positive infinity or negative infinity. So keeping track of signs is imperative here. So on the other hand, if we look at this thing right here, if you have a positive four on the top and you divide by something a little bit bigger than zero, that is, it's positive. You take a positive divided by a very small positive. That's going to give you a very huge positive. Thus, this is going to translate to mean for us, we're going to get positive infinity. So drawing our picture back on the screen, we see that as we approach x equals two from the right, that our function is in fact going up towards positive infinity. On the other hand, if we take the left handed limit, so we're going to take x as it approaches two from the left of three x minus two over x minus two, then plug in two here for the numerator, we can get away with a two, right? You're going to get three times two minus two on the bottom. You're going to get two minus minus two because again, in the numerator, you're going to end up with a four again. Technically speaking, it would be four minus those were we're approaching four from the left. So we're getting numbers that are a little bit, little bit smaller than four. But in terms of a sign, again, we want something that's positive. If you're just a teeny bit smaller than four, you're going to be positive. So that symbol is not necessary. But in the denominator, which went to zero, it's significant. If we take a number that's a little bit smaller than two and subtract from a two, we're going to get something that's a little bit smaller than zero. In particular, it's going to be a negative number. So if you take a positive four and divide it by a negative number, which has a really, really, really small absolute value, the number will have huge absolute value, but it'll be negative. And hence, this gives us negative infinity. So we see here that as you approach two from the left, you'll get negative infinity. As you approach two from the right, you're going to get positive infinity. And so the left limit will be positive infinity, excuse me, the left limit will be negative infinity. The right limit will be positive infinity. So to answer the original question about what happens to the limit as x approaches two of three x minus two over x minus two, we see that this limit does not exist because the left limit is negative infinity. The right limit is positive infinity. There's disagreement there. Let's try this again a little bit faster this time now that we've seen the technique. If we plug in x equals two, you're going to see the denominator. This is going to look like, well, in the numerator, you're going to get four again, excuse me, three times two, which is six minus two is four. But in the denominator, you're going to end up with a zero, a zero squared, mind you, which gives you just zero. So that's the form here. So again, we know there's a vertical asymptote at x equals two, but we need to learn more about it by looking at the left and the right hand limits. If we take the limit as x approaches two from the right, three x minus two over x minus two squared, we see that the top will be a four. Two, two plus minus two is going to be zero plus again. That's just like before you square it. So zero plus means we're approaching zero from the positive side. We're taking numbers that are very, very small, but still positive. So things just a little bit bigger than zero. Well, if you take something that's positive and you square, you're going to still get something that's positive. So you're going to get four over zero plus, which we interpret that as positive infinity. On the other hand, though, if we take the limit as x approaches two from the left of three x minus two over x minus two squared, we end up with four over zero minus like we did in the previous exercise. But zero minus means we're approaching zero from the left hand side from the negative side. If you take negative numbers and you square them, those are always positive. So zero minus squared is actually zero plus. And so this gives us positive infinity. So what this tells us about the graph is that our function here, the approach from the right is going towards infinity. The approach from the left is also going towards positive infinity. Therefore, the limit as x approaches two of three x minus two over x minus two squared, this will be positive infinity. All right, let's look at a final example of such a thing like this. You'll notice that I did switch up the numerator a little bit, we still have an x minus two on the bottom. So when you plug in x equals two, the bottom is going to look like a zero cube, which is equal to zero, of course. In the numerator, it's a little bit more complicated. You're going to have a negative, let's see, you're going to get a 12 minus a 14 plus a two. That's actually zero, right? You get zero over zero. That does not necessarily mean a vertical asymptote. It gives you an indeterminate form. It turns out anything could be happening at zero over zero. I mean, it's going to simplify the fraction. Really, what we are discovering here is that the numerator is divisible by x minus two. The denominator is obviously divisible by x minus two, because it's already factored, x minus two cubed, but we need to factor the numerator. So after some effort, you can discover that the numerator factors as three minus one, three x minus one times x minus two. Like so, for which then if you cancel the x minus two with one of the x minus two is on the bottom, you'll get a square there. We end up with the limit as x approaches two of negative three x minus one over x minus two squared, for which if we plug in our approach to two, we're going to get negative three times two minus one over, I'm going to put two to the plus minus minus two squared right here, because we have to pay attention. Are we approaching from the left or the right of two, because that gives us zero. Now in the numerator, you get three times two, which is six minus one, which is five, you're going to get a negative five on top. On the bottom, you're going to get zero plus or minus squared. But like we saw in the previous example, if you square positive, you get positive, you square negative, you get negative. And so in either case, you're going to get negative five over zero plus. So look at the signs there, the numerator is going to be negative as we get close to negative two, and the positive two, excuse me, and then the denominator is going to be positive as we get close to two. And so therefore this tells us that the limit is equal to negative infinity.